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3 years ago

6

As an electron moves through a region of space, its speed decreases from 6.03 × 106 m/s to 2.45 × 106 m/s. the electric force is

the only force acting on the electron. (a) did the electron move to a higher potential or a lower potential? (b) across what potential difference did the electron travel?

1 answer:

yuradex [85]3 years ago

7 0

Answer:

a) It moved to a lower potential

b) ΔФ = - 86.28 Volt

Explanation:

The energy of an charged particle and the electric potential are related by the potential al kinetic energies:

$\frac{\text{mv}^2}{2} = e\phi$

If we consider an electron:

m = 9.10938*10^-31 Kilogram

e = 1.60218*10^-19 Coulomb

And the potential diference may be calculed by:

$\Delta \phi =\frac{m \left(v_f^2-v_i^2\right)}{2 e}$

Replacing all the values we get:

ΔФ = - 86.28 (Kilogram Meter^2)/(Coulomb Second^2) = -86.28 Volts

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Four different resistors have various amounts of electric current flowing through them. Given the values of current I and resist

Artemon [7]

You did not provide the options. However, the options are

I = 6.0, R= 4.0 ohms

I = 9.0, R= 2.0ohms

I = 3.0, R= 2.0ohms

I = 8.0, R= 8.0 ohms

Answer:

The order of the resistors from the highest to the lowest is:

I = 8.0, R= 8.0 ohms

I = 6.0, R= 4.0 ohms

I = 9.0, R= 2.0ohms

I = 3.0, R= 2.0 ohms

Explanation:

ohm's law states that voltage across a conductor is directly proportional to the current flowing through it. V = IR

Based on this formula, the voltages in each of the resistors are calculated below from the highest to the lowest

For I = 8.0, R= 8.0 ohms

V = 8 * 8 =64 volts

For I = 6.0, R= 4.0 ohms

V = 6 * 4 =24 volts

For I = 9.0, R= 2.0 ohms

V = 9 * 2 =18 volts

For I = 3.0, R= 2.0 ohms

V = 3 * 2 =6 volts

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3 years ago

A baseball hits a car, breaking its window and triggering its alarm which sounds at a frequency of 1210 Hz. What frequency (in H

IRINA_888 [86]

Answer:

The frequency of sound heard by the boy is 1181 Hz.

Explanation:

Given that,

Frequency of sound from alarm $f_{0} = 1210\ Hz$

Speed = -8.25 m/s

Negative sign show the boy riding away from the car

Speed of sound = 343

We need to calculate the heard frequency

Using formula of frequency

$f = f_{0}(\dfrac{v+v_{0}}{v-v_{s}})$

Where, $f_{0}$ = frequency of source

$v_{0}$ = speed of observer

$v_{s}$ = speed of source

$v$ = speed of sound

Put the value into the formula

$f=1210\times\dfrac{343+(-8.25)}{343-0}$

here, source is at rest

$f=1180.8\ Hz$

$f=1181\ Hz$

Hence, The frequency of sound heard by the boy is 1181 Hz.

8 0

4 years ago

A weightlifter curls a 30 kg bar, raising it each time a distance of 0.60 m. How many times must he repeat this exercise to burn

Sholpan [36]

Answe

given,

mass of the bar, m = 30 Kg

distance of rise, h = 0.60 m

Assuming the efficiency = 25 %

energy from the pizza slice = 300 C = 1260 kJ

To consume Energy from the pizza bar is to be pulled several number of time.

( energy from pizza ) x (efficiency) = n m g h

n is the number of lift

( 1260 x 10³) x (0.25) = n x 30 x 9.8 x 0.6

$n = \dfrac{315\times 10^3}{176.4}$

n = 1786 times.

Weightlifter should lift bar 1786 times to burn off the energy.

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3 years ago

A tennis ball is released from a height of 4.0 m above the floor. After its third bounce off the floor, it reaches a height of 1

diamong [38]

Answer:

The percentage of its mechanical energy does the ball lose with each bounce is 23 %

Explanation:

Given data,

The tennis ball is released from the height, h = 4 m

After the third bounce it reaches height, h' = 183 cm

= 1.83 m

The total mechanical energy of the ball is equal to its maximum P.E

E = mgh

= 4 mg

At height h', the P.E becomes

E' = mgh'

= 1.83 mg

The percentage of change in energy the ball retains to its original energy,

$\Delta E\%=\frac{1.83mg}{4mg}\times100\%$

ΔE % = 45 %

The ball retains only the 45% of its original energy after 3 bounces.

Therefore, the energy retains in each bounce is

∛ (0.45) = 0.77

The ball retains only the 77% of its original energy.

The energy lost to the floor is,

E = 100 - 77

= 23 %

Hence, the percentage of its mechanical energy does the ball lose with each bounce is 23 %

5 0

4 years ago

A 2 kg blob of putty moving at 3 m/s slams into a 3 kg blob of putty at rest. Which blob has the most momentum prior to the coll

Ray Of Light [21]

initial momentum of 2 kg blob is given as

$P_1 = m_1v_1$

here we have

$m_1 = 2kg$

$v_1 = 3m/s$

$P_1 = 3(2) = 6 kg m/s$

initial momentum of 3 kg blob is given as

$P_2 = m_2v_2$

here we have

$m_2 = 3kg$

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$P_2 = 3(0) = 0 kg m/s$

So initially 2 kg Blob has most momentum before they collide

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3 years ago